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4 years ago

What is the mass of 3.8 × 1024 atoms of argon (Ar)?

Chemistry

1 answer:

gulaghasi [49]4 years ago

6 0

40 g Ar.......6.023*10^23 atoms
x g Ar.......3.8*10^24 atoms
x=40*3.8*10^24/(6.023*10^23)=252,365g Ar

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Answer:

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Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

$r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm$

Density is equal to mass per unit volume.

$d=\dfrac{m}{V}$

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V is the volume of the cylinder

$V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3$

So,

$m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g$

So, the mass of the cylinder is 1.51 grams.

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- Source: google

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